## Precalculus (6th Edition) Blitzer

The value of $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}$ is $\frac{1}{6}$.
Consider the function $f\left( x \right)=\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}$. Simplify the above function, Multiplying the numerator and the denominator of the function by $\sqrt{{{x}^{2}}+9}+3$, So, \begin{align} & f\left( x \right)=\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}\left( \frac{\sqrt{{{x}^{2}}+9}+3}{\sqrt{{{x}^{2}}+9}+3} \right) \\ & =\frac{{{x}^{2}}+9-9}{{{x}^{2}}\left( \sqrt{{{x}^{2}}+9}+3 \right)} \\ & =\frac{{{x}^{2}}}{{{x}^{2}}\left( \sqrt{{{x}^{2}}+9}+3 \right)} \\ & =\frac{1}{\sqrt{{{x}^{2}}+9}+3} \end{align} The function $g\left( x \right)={{x}^{2}}+9$ is a polynomial. Now, find the value of $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}$, \begin{align} & \underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{\sqrt{{{x}^{2}}+9}+3} \\ & =\frac{1}{\sqrt{\underset{x\to 0}{\mathop{\lim }}\,{{x}^{2}}+9}+3} \\ & =\frac{1}{\sqrt{0+9}+3} \\ & =\frac{1}{\sqrt{9}+3} \end{align} \begin{align} & =\frac{1}{3+3} \\ & =\frac{1}{6} \\ \end{align} Thus, the value of $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}$ is $\frac{1}{6}$.